$n_C=n_{CO_2}=\dfrac{7,04}{44}=0,16(mol)$
$n_H=2n_{H_2O}=0,128.2=0,256(mol)$
Gọi CTPT este X là $C_xH_yO_2$
$\Rightarrow \dfrac{x}{y}=\dfrac{0,16}{0,256}=\dfrac{5}{8}$
Vậy X là $C_5H_8O_2$
Ancol Y: $C_nH_{2n+2-2k}O$
$C_nH_{2n+2-2k}O + \dfrac{3n-k}{2}O_2\to nCO_2+(n+1-k)H_2O$
$\Rightarrow \dfrac{3n-k}{2}=3$
$\Leftrightarrow 3n-k=6$
$\Rightarrow k=0; n=2 (C_2H_5OH)$
Vậy este X là $CH_2=CHCOOC_2H_5$ (etyl acrylat)